Question

Resolva a equação abaixo: log 10(4x-2)=log 10²- log 10 (2x-1)​

Answer 1

$\log{(4x-2)}=\log{10^2}-\log{(2x-1)}\ \therefore$

$\log{(4x-2)}=\log{\bigg(\dfrac{100}{2x-1}\bigg)}\ \therefore\ 2(2x-1)=\dfrac{100}{2x-1}\ \therefore$

$(2x-1)(2x-1)=50\ \therefore\ 4x^2-4x+1=50\ \therefore$

$4x^2-4x-49=0\ \to\ a=4,\ b=-4,\ c=-49$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(-49)}}{2(4)}=$

$\dfrac{4\pm\sqrt{800}}{8}=\dfrac{4\pm20\sqrt{2}}{8}=\dfrac{1\pm5\sqrt{2}}{2}$

$2x-1>0\ \therefore\ x>\dfrac{1}{2}\ \to\ \boxed{x=\dfrac{1+5\sqrt{2}}{2}}$